Mean
"\\mu=\\dfrac{2+4+6+8+10+12}{6}=7"Variance
"\\sigma^2=\\dfrac{1}{6}\\big((2-7)^2+(4-7)^2+(6-7)^2""+(8-7)^2+(10-7)^2+(12-7)^2\\big)""=\\dfrac{70}{6}=\\dfrac{35}{3}"
Standard deviation
"\\sigma=\\sqrt{\\sigma^2}=\\sqrt{\\dfrac{35}{3}}\\approx\t3.415650"
We have population values "2,4,6,8,10,12" population size "N=6" and sample size "n=3." Thus, the number of possible samples which can be drawn without replacement is
"\\dbinom{6}{3}=\\dfrac{6!}{3!(6-3)!}=20""\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c}\n Sample & Sample & Sample \\ mean \\\\\n No. & values & (\\bar{X}) \\\\ \\hline\n 1 & 2,4,6 & 4 \\\\\n \\hdashline\n 2 & 2,4,8 & 14\/3 \\\\\n \\hdashline\n 3 & 2,4,10 & 16\/3 \\\\\n \\hdashline\n 4 & 2,4,12 & 6 \\\\\n \\hdashline\n 5 & 2,6,8 & 16\/3 \\\\\n \\hdashline\n 6 & 2,6,10 & 6 \\\\\n \\hdashline\n 7 & 2,6,12 & 20\/3 \\\\\n \\hdashline\n 8 & 2,8,10 & 20\/3 \\\\\n \\hdashline\n 9 & 2,8,12 & 22\/3 \\\\\n \\hdashline\n 10 & 2,10,12 & 8 \\\\ \n \\hdashline\n 11 & 4,6,8 & 6 \\\\ \n \\hdashline\n 12 & 4,6,10 & 20\/3 \\\\ \n \\hdashline\n 13 & 4,6,12 & 22\/3 \\\\ \n \\hdashline\n 14 & 4,8,10 & 22\/3 \\\\ \n \\hdashline\n 15 & 4,8,12 & 8 \\\\ \n \\hdashline\n 16 & 4,10,12 & 26\/3 \\\\ \n \\hdashline\n 17 & 6,8,10 & 8 \\\\ \n \\hdashline\n 18 & 6,8,12 & 26\/3 \\\\ \n \\hdashline\n 19 & 6,10,12 & 28\/3 \\\\ \n \\hdashline\n 20 & 8,10,12 & 10 \\\\ \n \\hdashline\n\\end{array}"
The sampling distribution of the sample means.
"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n \\bar{X} & f & f(\\bar{X}) & \\bar{X}f(\\bar{X})& \\bar{X}^2f(\\bar{X}) \\\\ \\hline\n 4 & 1& 1\/20 & 12\/60 & 144\/180 \\\\\n \\hdashline\n 14\/3 & 1 & 1\/20 & 14\/60 & 196\/180 \\\\\n \\hdashline\n 16\/3 & 2 & 2\/20 & 32\/60 & 512\/180 \\\\\n \\hdashline\n 6 & 3 & 3\/20 & 54\/60 & 972\/180\\\\\n \\hdashline\n 20\/3 & 3 & 3\/20 & 60\/60 & 1200\/180 \\\\\n \\hdashline\n 22\/3 & 3 & 3\/20 & 66\/60 & 1452\/180 \\\\\n \\hdashline\n 8 & 3 & 3\/20 & 72\/60 & 1728\/180 \\\\\n \\hdashline\n 26\/3 & 2 & 2\/20 & 52\/60 & 1352\/180 \\\\\n \\hdashline\n 28\/3 & 1 & 1\/20 & 28\/60 & 784\/180 \\\\\n \\hdashline\n 10 & 1 & 1\/20 & 30\/60 & 900\/180 \\\\\n \\hdashline\n Total & 20 & 1 & 420\/60 & 9240\/180 \\\\ \\hline\n\\end{array}"
"E(\\bar{X})=\\sum\\bar{X}f(\\bar{X})=\\dfrac{420}{60}=7"The mean of the sampling distribution of the sample means is equal to the
the mean of the population.
"E(\\bar{X})=\\mu_{\\bar{x}}=7=\\mu"
"Var(\\bar{X})=\\sum\\bar{X}^2f(\\bar{X})-(\\sum\\bar{X}f(\\bar{X}))^2""=\\dfrac{9240}{180}-(7)^2=\\dfrac{7}{3}""\\sigma_{\\bar{X}}=\\sqrt{Var(\\bar{X})}=\\sqrt{\\dfrac{7}{3}}\\approx1.527525"Verification:
"Var(\\bar{X})=\\dfrac{\\sigma^2}{n}(\\dfrac{N-n}{N-1})=\\dfrac{35}{3(3)}(\\dfrac{6-3}{6-1})""=\\dfrac{7}{3}, True"
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